1. ## Evaluating integrals

I have a final monday and have been working on this integral and cant figure it out.

evaluate integral 1/(e^(3x)-e^(x))

2. Set $u=e^x$. Then you get a rational function $\frac{1}{u(u-1)(u+1)}$. Rewrite this rational function as a sum of rational functions $\frac{A}{u}+\frac{B}{u+1}+\frac{C}{u-1}$, for some constants $A,B,C$. Then integrate.

3. All you need to do is to see that $e^{3x} = (e^{x})^{3}$

Sub in $u = e^{x} , du = e^{x}$

Edit

vemrygh beat me to it.

4. $\frac{1}{u(u-1)(u+1)}=\frac{1}{u\left( u^{2}-1 \right)}=\frac{u^{2}-\left( u^{2}-1 \right)}{u\left( u^{2}-1 \right)}=\frac{u}{u^{2}-1}-\frac{1}{u}.$

5. $\frac{1}{e^{3x}-e^{x}} =$
$\frac{-2e^{2x} + 2 + e^{3x} + e^{2x} - e^{3x} + e^{2x}}{2(e^{3x}-e^{x})}=$
$\frac{2(e^x-1)(e^x+1)+(e^x)(e^x+1))(e^x)-(e^x)(e^x-1))(e^x)}{2*(e^x)(e^x-1)(e^x+1)}=
$

$-\frac{2(e^x-1)(e^x +1)}{2(e^x)(e^x-1)(e^x +1)} +\frac{(e^x)(e^x +1)(e^x)}{2(e^x)(e^x-1)(e^x +1)} -\frac{(e^x)(e^x-1)(e^x)}{2(e^x)(e^x-1)(e^x +1)}=$
$-\frac{1}{e^x} + \frac{e^x(e^x+1)}{2(e^x-1)(e^x+1)} - \frac{e^x(e^x-1)}{2(e^x+1)(e^x-1)}=$
$-\frac{1}{e^x} + \frac{e^x}{2(e^x-1)} - \frac{e^x}{2(e^x+1)}$

which can easily be integrated to:
$\frac{1}{e^x} + \frac{1}{2}ln(e^x-1)- \frac{1}{2}ln(e^x+1)$